JEE Advanced 2026 — Paper 2 — CHEM-I solution

JEE Advanced2026Paper 2CHEM-I

The correct order of ONO bond angle in the given species is:

A
NO $_{2}^{+}$ $<$ NO $_{2}$ $<$ NO $_{3}^{-}$ $<$ NO $_{2}^{-}$
B
NO $_{2}^{-}$ $<$ NO $_{3}^{-}$ $<$ NO $_{2}$ $<$ NO $_{2}^{+}$
C
NO $_{3}^{-}$ $<$ NO $_{2}^{-}$ $<$ NO $_{2}$ $<$ NO $_{2}^{+}$
D
NO $_{2}^{-}$ $<$ NO $_{3}^{-}$ $<$ NO $_{2}^{+}$ $<$ NO $_{2}$

Solution

VSEPR analysis of each species (focus on the angle at N between the two given oxygens).

NO $_{2}^{+}$ : linear, sp hybridized at N, no lone pair on N. ONO angle $= 18 0^{\circ}$ .

NO $_{2}$ (neutral): sp $^{2}$ at N with one unpaired electron (single non-bonding electron, less repulsive than a lone pair). ONO angle $\approx 13 4^{\circ}$ .

NO $_{3}^{-}$ : trigonal planar, sp $^{2}$ at N, three equivalent N-O bonds, no lone pair on N. ONO angle $= 12 0^{\circ}$ .

NO $_{2}^{-}$ : bent, sp $^{2}$ at N with one full lone pair which gives stronger lp-bp repulsion than the single electron in neutral NO $_{2}$ . ONO angle $\approx 11 5^{\circ}$ .

Order: $11 5^{\circ} < 12 0^{\circ} < 13 4^{\circ} < 18 0^{\circ}$ , i.e. NO $_{2}^{-}$ $<$ NO $_{3}^{-}$ $<$ NO $_{2}$ $<$ NO $_{2}^{+}$ , matching option (B).

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